In rectangle $ADEH$, points $B$ and  $C$  trisect $\overline{AD}$, and points $G$ and  $F$ trisect $\overline{HE}$. In addition,  $AH=AC=2$. What is the area of  quadrilateral $WXYZ$ shown in the figure?

[asy]
unitsize(1cm);
pair A,B,C,D,I,F,G,H,U,Z,Y,X;
A=(0,0);
B=(1,0);
C=(2,0);
D=(3,0);
I=(3,2);
F=(2,2);
G=(1,2);
H=(0,2);
U=(1.5,1.5);
Z=(2,1);
Y=(1.5,0.5);
X=(1,1);
draw(A--D--I--H--cycle,linewidth(0.7));
draw(H--C,linewidth(0.7));
draw(G--D,linewidth(0.7));
draw(I--B,linewidth(0.7));
draw(A--F,linewidth(0.7));
label("$A$",A,SW);
label("$B$",B,S);
label("$C$",C,S);
label("$D$",D,SE);
label("$E$",I,NE);
label("$F$",F,N);
label("$G$",G,N);
label("$H$",H,NW);
label("$W$",U,N);
label("$X$",X,W);
label("$Y$",Y,S);
label("$Z$",Z,E);
[/asy]
Answer: First note that since points $B$ and  $C$  trisect $\overline{AD}$, and points $G$ and  $F$ trisect $\overline{HE}$, we have $HG = GF = FE = AB = BC = CD = 1$. Also,  $\overline{HG}$ is parallel to $\overline{CD}$ and $HG = CD$, so $CDGH$ is a parallelogram. Similarly,  $\overline{AB}$ is parallel to $\overline{FE}$ and $AB = FE$, so $ABEF$ is a parallelogram. As a consequence,  $WXYZ$ is a parallelogram, and since $HG = CD = AB = FE$, it  is a rhombus.

[asy]
unitsize(1cm);
pair A,B,C,D,I,F,G,H,U,Z,Y,X;
A=(0,0);
B=(1,0);
C=(2,0);
D=(3,0);
I=(3,2);
F=(2,2);
G=(1,2);
H=(0,2);
U=(1.5,1.5);
Z=(2,1);
Y=(1.5,0.5);
X=(1,1);
draw(A--D--I--H--cycle,linewidth(0.7));
draw(H--C,linewidth(0.7));
draw(G--D,linewidth(0.7));
draw(I--B,linewidth(0.7));
draw(A--F,linewidth(0.7));
label("$A$",A,SW);
label("$B$",B,S);
label("$C$",C,S);
label("$D$",D,SE);
label("$E$",I,NE);
label("$F$",F,N);
label("$G$",G,N);
label("$H$",H,NW);
label("$W$",U,N);
label("$X$",X,W);
label("$Y$",Y,S);
label("$Z$",Z,E);
draw(F--C,linewidth(0.5));
[/asy]

Since $AH = AC = 2$, the rectangle $ACFH$ is a square of side length 2. Its diagonals $\overline{AF}$ and $\overline{CH}$ have length $2\sqrt{2}$ and form a right angle at $X$. As a consequence, $WXYZ$ is a square. In isosceles $\triangle HXF$ we have $HX = XF = \sqrt{2}$. In addition, $HG = \frac{1}{2}HF$. So $XW = \frac{1}{2}XF = \frac{1}{2}\sqrt{2}$, and the square $WXYZ$ has area $XW^2 = \boxed{\frac{1}{2}}$.